Hi Dominika,
The statement we want to prove is the following:
Suppose that $f$ is differentiable on it's domain $(a,b)$. Then $f$ is convex if and only if $(a,b)$ is convex and $f(x)\geq f(c)+f^{\prime}(c)(x-c)$ for all $x,c\in (a,b)$.
We're going to assume (without proof) the fact that an interval is convex (i.e. for any $x,y\in (a,b)$, we have $ty+(1-t)x\in (a,b)$ for any $0\leq t\leq 1$).
We proceed with the proof as follows.
Proof: Suppose $f$ is convex on $(a,b)$. Then for any $t\in[0,1]$ and $x,c\in(a,b)$, $tx + (1-t)c = c + t(x-c) \in (a,b)$. Furthermore, by
definition of convexity of $f$, we have that
\[f(tx+(1-t)c) = f(c+t(x-c)) \leq tf(x) + (1-t)f(c).\]
Dividing both sides by $t$ and then solving the inequality for $f(x)$ gives us
\[f(x) \geq f(c) + \frac{f(c+t(x-c))-f(c)}{t}\]
Taking the limit as $t\to 0$ gives us
\[f(x)\geq f(c) + f^{\prime}(c)(x-c),\tag{1}\]
which is the desired result ($f(x)\geq\text{tangent line}$).
Conversely, suppose that $f$ lies above any tangent line over the interval $(a,b)$. Choose $x,c\in(a,b)$ such that $x\neq c$. Now, for any $t\in [0,1]$, define $z=tx + (1-t)c= c+t(x-c)$. We can see from here that $\color{red}{x-z}= \color{red}{(1-t)(x-c)}$ and $\color{blue}{c-z}=\color{blue}{-t(x-c)}$. Applying $(1)$ twice gives us
\[f(x)\geq f(z) + f^{\prime}(z)(x-z)\tag{2}\]
and
\[f(c)\geq f(z)+f^{\prime}(z)(c-z).\tag{3}\]
If we now multiply $(2)$ by $t$ and $(3)$ by $(1-t)$ and add them together, we see that
\[\begin{aligned} tf(x)+(1-t)f(c) &\geq tf(z)+(1-t)f(z) + tf^{\prime}(z)(\color{red}{x-z}) + (1-t)f^{\prime}(z)(\color{blue}{c-z})\\ &= f(z) + tf^{\prime}(z)\color{red}{(1-t)(x-c)} + (1-t)f^{\prime}(z)(\color{blue}{-t(x-c)})\\ &= f(z),\end{aligned}\]
which is the condition of convexity. Therefore, $f$ is convex.$\hspace{.25in}\blacksquare$
I hope this makes sense! (Smile)